- [Instructor] Hey folks.

In this video I wanna show you something pretty interesting

about these Lagrange multipliers that we've been studying.

So the first portion I'm just gonna kind of get this set up,

which is a lot of review from what we've seen already,

but I think you're really gonna like where this is going

in the end.

So one of the examples I showed, and I think this is a

pretty nice prototypical example

for constrained optimization problems,

is that you're running a company

and you have some kind of revenue function

that's dependent on various choices

you make in running the company,

and I think I said the number of hours of labor

you employ and the number of tons of steel you use,

you know, if you were manufacturing something metallic.

And, you know, this might be modeled as some multivariable

function of h and s, right now we don't really care

about the specifics.

And you're trying to maximize this, right,

that's kind of the whole point of this unit

that we've been doing, is that you're trying to maximize

some function, but you have a constraint.

This is the real world, you can't just spend infinite money,

you have some kind of budget, some sort of amount of money

you spend as a function of those same choices you make,

the hours of labor you employ, the tons of steel you use.

And this, again, it's gonna equal some

multivariable function that tells you, you know,

how much money you spend for a given amount of hours

and given number of tons of steel.

And you set this equal to some constant,

this tells you the amount of money you're willing to spend.

And our goal has been to maximize some function,

subject to a constraint like this.

And the mental model you have in mind

is that you're looking in the input space,

inside the x-y plane,

or I guess really, it's the h-s plane in this case, right.

Your inputs are h and s, and points in this plane

tell you possible choices you can make

for hours of labor and tons of steel.

And you think of this budget

as some kind of curve in that plane, right.

All the sets of h and s that equal $10,000

is gonna give you some kind of curve.

And the core value we care about is that,

when you maximize this revenue,

you know, when you set it equal to a constant

I'm gonna call M star, that's like the maximum possible

revenue, that's gonna give you a contour

that's just barely tangent

to the constraint curve.

And if that seems unfamiliar,

definitely take a look at the videos preceding this one.

But, just to kind of continue the review,

this gave us the really nice property

that you look at the gradient vector

for the thing you're trying to maximize, R,

and that's gonna be proportional to the gradient vector

for the constraint function, for this B, so gradient of B.

And this is because gradients are perpendicular

to contour lines.

Again, this should feel mostly like review at this point.

So the core idea was that we take this gradient of R,

and then make it proportional, with some kind of

proportionality constant lambda,

to the gradient of B, to the gradient

of the constraint function.

And up till this point, this value lambda

has been wholly uninteresting.

It's just been a proportionality constant, right,

because we couldn't guarantee that the gradient of R

is equal to the gradient of B.

All we care about is that they're pointing

in the same direction.

So we just had this constant sitting here,

and all we really said is make sure it's not zero.

But here, we're gonna get to where this little guy

actually matters.

So, if you'll remember, in the last video,

I introduced this function called the Lagrangian,

the Lagrangian.

And it takes in multiple inputs,

they'll be the same inputs that you have

for your budget function and your revenue function,

or more generally, the constraint

and the thing you're trying to maximize.

It takes in those same variables,

but, also, as another one of its inputs,

it takes in lambda.

So, it is a higher-dimensional function

than both of these two, because we've got this extra lambda.

And the way it's defined looks a little strange at first,

it just seems kind of like this random

hodgepodge of functions that we're putting together.

But, last time, I kind of walked through

why this makes sense.

You take the thing you're trying to maximize,

and you subtract off this lambda,

multiplied by the constraint function,

which is B of those inputs,

minus, and then whatever this constant is here, right.

I'm gonna give it a name, I'm gonna call this constant

lowercase b.

So maybe we're thinking of it as $10,000,

but it's whatever your actual budget is.

So we think of that, and I'm just gonna

emphasize here that that's a constant, right,

that this b is being treated as a constant right now.

You know, we're thinking of h and s and lambda

all as these variables, and this gives us

some multivariable function.

And if you'll remember from the last video,

the reason for defining this function

is it gives us a really nice compact way to solve

the constraint optimization problem.

You set the gradient of L equal to zero,

or really the zero vector, right, it'll be a vector

with three components here.

And when you do that, you'll find some solution, right,

you'll find some solution, which I'll call h star,

s star, and lambda, here I'll give it that green

lambda color, lambda star.

You'll find some value that, when you input this

into the function, the gradient will equal zero.

And, of course, you might find multiple of these, right,

you might find multiple solutions to this problem,

but what you do is, for each one of them you're gonna take

a look at h star and s star, then you're gonna plug those

into the revenue function, or the thing that you're trying

to maximize.

And, typically, you only get a handful, you get a number,

then you can actually plug each one of them into

the revenue function, and you'll just check which one

of them makes this function the highest.

And whatever the highest value this function can achieve,

that is M star, that is the maximum possible revenue,

subject to this budget.

But it's interesting that when you solve this, you get some

specific value of lambda, right, there's a specific

lambda star that will be associated with the solution.

And, like I said, this turns out not to just be

some dummy variable.

It's gonna carry information about how much we can increase

the revenue if we increase that budget.

And, here, let me show you what I mean.

So we've got this M star, and I'll just write it again,

M star here.

And what that equals, I'm saying that's the maximum

possible revenue.

So that's gonna be the revenue when you evaluate it at

h star,

h star and s star.

And h star and s star, they are whatever the solution

to this gradient of the Lagrangian equals zero equation is.

You set this multivariable function equal to the

zero vector, you solve when each of its partial derivatives

equal zero, and you'll get some kind of solution.

So when you plug that solution in the revenue,

that gives you the maximum possible revenue.

But what we could do is consider this as a function

of the budget.

Now, this is the kind of thing that looks a little bit wacky

if you're just looking at the formulas.

But if you actually think about what it means

in this context of kind of a revenue and a budget,

I think it's actually pretty sensible,

where, really, if we consider this b no longer to be

constant but something that you could change, right,

you're wondering, well what if I had a $20,000 budget,

or what if I had a $15,000 budget?

You wanna ask the question, what happens as you change

this b.

Well, the maximizing value, h star and s star,

each one of those guys is gonna depend on b, right.

As you change what this constant is, it's gonna change

the values at which the gradient of the Lagrangian

equals zero.

So, I'm gonna rewrite this function as

the revenue evaluated at h star and s star,

but now I'm gonna consider that h star and s star

each as functions of b, right,

because they depend on it in some way.

As you change b, it changes the solution to this problem

It's very implicit and it's kind of hard to think about.

It's hard to think, okay, as I change this b, how much

does that change h star.

Well that depends on what the, you know,

what the definition of R was and everything there.

But, in principle, in this context, I think it's quite

intuitive.

You have a maximum possible revenue,

and that depends on what your budget is.

So, what turns out to be a beautiful, absolutely beautiful

magical fact

is that this lambda star

is equal to the derivative of M star,

the derivative of this maximum possible revenue

with respect to b, with respect to the budget.

And let me just show you what that actually means, right.

So if, for example, let's say you did

all of your calculations and it turned out that lambda star

was equal to 2.3.

You know, previously that just seemed like some dummy number

that you ignore, and you just look at whatever the

associated values here are.

But if you plug this in the computer and you see lambda star

equals 2.3, what that means is, for a tiny change in budget,

like let's say your budget increases from 10,000

to 10,001, it goes up to $10,001,

you increase your budget by just a little bit, a little db.

Then the ratio of the change in the maximizing revenue

to that db is about 2.3.

So what that would mean is, increasing your budget

by a dollar is gonna increase M star,

over here it would mean that M star increases

by about, you know, $2.30

for every dollar that you increase your budget.

And that's information you'd wanna know, right?

If you see that this lambda star is a number

bigger than one, you'd say, hey, maybe we should increase

our budget.

We increase it from $10,000 to 10,001 and we're making

more money.

So, maybe, as long as lambda star is greater than one,

you should keep doing whatever it takes to increase

that budget.

So this fact is quite surprising, I think,

and it seems like it totally comes out of nowhere.

So what I'm gonna do in the next video is prove this to you,

is prove why this is true, why this lambda star value

happens to be the rate of change for the maximum value

of the thing we're trying to maximize

with respect to this constant, with respect to whatever

constant you set your constraint function equal to.

For right now, though, I just want you to kind of

try to sit back and digest what this means in the context

of this specific economic example.

And, even if you never looked into the proof

and never understood it there,

I think this is an interesting and even useful

tidbit of knowledge to have about Lagrange multipliers.

So with that, I'll see you in the next video.